How many atoms are 1.00?10^-9 g of lead?

Dear student

Molar mass of lead = 207.2
Given mass of lead = $1 \times {10}^{-9 } g$
Avogadro constant =  $6.022 \times {10}^{23}$

$$\begin{gathered} No. of atoms = \frac{Given mass}{Molar mass} \times Avogadro cons\tan t \\ = \frac{ 1.0 \times {10}^{-9}}{207.2} 6.022 \times {10}^{23} \\ = 2.90 \times {10}^{12 } atoms \end{gathered}$$


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